the process of diagonal izing matrix is
closely related to finding the eigen
values and the eigenvectors of a given
matrix the crux of of this idea is
captured in the following equation in
diagonalizing a matrix will say a
assuming its Squared's and by n like to
do is write that matrix in this
particular factored form that is as the
product of three matrices p and p
inverse where specifically d for
diagonal here is a diagonal matrix
consisting of the eigen values of a and
P is the eigen vector matrix with
respect to a in other words we write the
eigenvectors of a to construct p in
column form if possible we'd like to
given a matrix a diagonalize it meaning
write it in this particular factored
form now use the caveat if possible for
the reason that not every matrix can be
written in this particular factored form
so it's not possible to diagonalize
every single matrix what we need to be
able to diagonal I as a matrix is
essentially we need enough eigen vectors
to fill in the matrix P so two diagonal
eyes a matrix therefore we need n will
say distinct again it's an n-by-n matrix
and distinct eigen vectors that is an
important note here so note we'll say a
is diagonalizable if it has n distinct
eigen vectors and the reason we need
that criterion met is so that well if we
have n distinct eigen vectors I have
sort of a sufficient number to fill in
for P
and by the way that matrix will be
invertible as a theorem kind of leftover
from linear algebra guarantees the
existence of P inverses well when I fill
in the eigen values for the diagonal
matrix D and then correspondingly right
in the eigenvectors in column four min p
I need to maintain the same order for I
ghen dies and eigenvectors as we'll see
so there needs to be a correspondence
with for instance the first eigenvalue
listed in d needs to correspond with the
eigen first eigenvector listed in matrix
p so two diagonal eyes a matrix in other
words put in this particular factored
form I need first and foremost the
eigenvalues and the eigenvectors of that
matrix so I'll say well begin our method
here for diagonalizing a matrix with
finding once more our lambdas and their
associated I ghen vectors well we've
done that previously let's compute the
eigenvalues and eigenvectors for a new
matrix will make it a two by two matrix
so let's consider the matrix this time a
and let's work with the matrix as
follows 0121 so we've first off light to
know if it's diagonalizable in other
words other sufficient number of
distinct eigen vectors in this case we
would need two distinct eigen vectors to
properly diagonalize the matrix because
it's a two by two matrix we begin to
find by finding the eigen values and
then the associated I get vectors we get
begin by solving the characteristic
equation associated with that matrix
namely this one a minus lambda times the
identity we set that equal to zero so we
get on the left hand side it's called
the characteristic polynomial so let's
do this in a reasonable number of steps
here together so we have the determinant
of the two-by-two matrix a where I
essentially subtract lambda from the
main diagonal so that then leaves me
with lambda negative lambda excuse me 1
2 & 1 minus lambda for a minus lambda
will take the determinant of that
set it equal to zero so once again the
determinant of a two-by-two matrix is
computed ad minus BC so the resultant is
we have negative lambda times the
quantity 1 minus lambda minus BC here
minus 2 equals 0 so let's go ahead and
solve that quadratic equation in lambdas
efficiently as possible so I'll
distribute the negative lambda across
the binomial term here so i have
negative lambda negative lambda times
negative lambda is positive lambda
squared minus 2 equals 0 you could solve
this using the quadratic formula of
course always here though fortunately
enough this factors i'm just going to
rearrange things so it's a little more
sort of conventional looking i'll write
it in descending order with the power so
lambda squared minus lambda minus 2
equals 0 it's again that fortunately
enough factors for us we can write this
then as lambda minus 2 times the
quantity lambda plus 1 equals 0 so this
implies either lambda equals 2 or lambda
equals negative 1 so sure enough those
are the eigen values associated with
that particular matrix a so now that we
have those eigenvalues as 2 and negative
1 for the respective matrix a we can
then go ahead and as before compute the
eigenvectors associated with each of
those eigenvalues now I've gone ahead
and done that beforehand just to save a
little bit of time but you should check
of course that you can do this following
the procedures in the previous sections
so a representative eigenvector
associated with eigenvalue 2 is the
vector 1 2 and i'll label that v sub 2
to indicate that is the eigenvector
associated with eigenvalue to know on
the other hand one possible
representative I ghen vector associated
with the value negative one eigenvalue
negative one I can all use the subscript
negative one here just to be clear is
the vector 1 negative 1 so here then our
objective is to diagonalize the matrix a
if possible and indeed this matrix is
diagonalizable namely because we have
two distinct eigen vectors associated
with a let's go ahead and write out that
factorization and then confirm through
matrix multiplication that it is in fact
valid it does in fact equal a so I write
a as the matrix P again that's my matrix
of eigenvectors written in column form
so i'll just write it in the same order
i'll say one two and one negative one
there is my matrix p and then
importantly i right the diagonalized
matrix or the diagonal matrix here in
terms of the eigenvalues in the same
order in which i listed these
eigenvectors so in other words I have to
list the eigenvalue to first and that
goes on the diagonal component there and
then in the second column I list the the
eigenvalue associate with the second
listed eigen vector which is of course
negative one so there is my diagonal
matrix D again maintaining the same
order of the eigenvectors listed in p
and then lastly i compute the inverse of
p now I've done that for us beforehand
and you should check that indeed this
matrix P inverse when I multiply by P
does end up equaling the identity matrix
okay so there is my P inverse matrix now
only I want to go ahead and just to be
totally thorough here multiply those
matrices out and confirm that our result
is equal day okay so I've gone ahead and
multiply doubt those matrices you of
course want to check on your own that
you can do this successfully I'll remind
you that matrix multiplication is
associative so I can group the
multiplications however I like i've
multiplied first the the first two
matrices together that results in the
matrix on the left I then subsequently x
p inverse and sure enough i'm back to
the original matrix a so indeed we have
a factorization of a specifically we
have diagonalized the matrix a
you